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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

Note on group irregularity strength of disconnected graphs

Marcin Anholcer / Sylwia Cichacz / Rafał Jura / Antoni Marczyk
Published Online: 2018-03-02 | DOI: https://doi.org/10.1515/math-2018-0017

Abstract

We investigate the group irregularity strength (sg(G)) of graphs, i.e. the smallest value of s such that taking any Abelian group 𝓖 of order s, there exists a function f : E(G) → 𝓖 such that the sums of edge labels at every vertex are distinct. So far it was not known if sg(G) is finite for disconnected graphs. In the paper we present some upper bound for all graphs. Moreover we give the exact values and bounds on sg(G) for disconnected graphs without a star as a component.

Keywords: Irregularity strength; Graph weighting; Graph labeling; Abelian group

MSC 2010: 05C15; 05C78

1 Introduction

It is a well known fact that in any simple graph G there are at least two vertices of the same degree. The situation changes if we consider an edge labeling f : E(G) → {1, …, s} and calculate weighted degree (or weight) of each vertex x as the sum of labels of all the edges incident to x. The labeling f is called irregular if the weighted degrees of all the vertices are distinct. The smallest value of s that allows some irregular labeling is called irregularity strength of G and denoted by s(G).

The problem of finding s(G) was introduced by Chartrand et al. in [5] and investigated by numerous authors [1, 2, 7, 14, 16]. Best published general result due to Kalkowski et al. (see [12]) is s(G) ≤ 6n/δ. It was recently improved by Majerski and Przybyło ([15]) for dense graphs of sufficiently large order (s(G) ≤ (4+o(1))n/δ+4 in this case).

Fujie-Okamoto, Jones, Kolasinski and Zhang combined the concepts of graceful labeling and modular edge coloring into labeling called a modular edge-graceful labeling ([8, 10, 11]). They defined the modular edge-gracefulness of graphs as the smallest integer k(G) = kn for which there exists an edge labeling f : E(G) → ℤk such that the induced vertex labeling f : V (G) → ℤk defined by f(u)=vN(u)f(uv)modk

is one-to-one.

Assume 𝓖 is an Abelian group of order mn with the operation denoted by + and identity element 0. For convenience we will write ka to denote a+a+…+a (where element a appears k times), −a to denote the inverse of a and we will use ab instead of a+(−b). Moreover, the notation ∑aS a will be used as a short form for a1+a2+a3+…, where a1, a2, a3, … are all the elements of the set S. Recall that any group element ι ∈ 𝓖 of order 2 (i.e., ι ≠ 0 such that 2ι = 0) is called involution.

The order of an element a ≠ 0 is the smallest r such that ra = 0. It is well-known by Lagrange Theorem that r divides |𝓖| [9]. Therefore every group of odd order has no involution.

We consider edge labeling f : E(G) → 𝓖 leading us to the weighted degrees defined as the sums (in 𝓖) : w(v)=vN(u)f(uv)

The concept of 𝓖-irregular labeling is a generalization of modular edge-graceful labeling. In both cases the labeling f is called 𝓖-irregular if all the weighted degrees are distinct. However, the group irregularity strength of G, denoted sg(G), is the smallest integer s such that for every Abelian group 𝓖 of order s there exists 𝓖-irregular labeling f of G. Thus the following observation is true.

Observation 1.1

([3]). For every graph G with no component of order less than 3, k(G) ≤ sg(G).

The following theorem, determining the value of sg(G) for every connected graph G of order n ≥ 3, was proved by Anholcer, Cichacz and Milanič [4].

Theorem 1.2

([4]). Let G be an arbitrary connected graph of order n ≥ 3. Then sg(G)=n+2,ifGK1,32q+12forsomeintegerq1,n+1,ifn2mod4GK1,32q+12foranyintegerq1,n,otherwise.

In [10] it was proved in turn that for every connected graph G of order n ≥ 3 k(G)=n,if n2 mod 4,n+1,if n2 mod 4.

In order to distinguish n vertices in arbitrary (not necessarily connected) graph we need at least n distinct elements of 𝓖. However, n elements are not always enough, as the following lemma shows.

Lemma 1.3

([3]). Let G be a graph of order n. If n ≡ 2 mod 4, then there is no 𝓖-irregular labeling of G for any Abelian group 𝓖 of order n.

Anholcer and Cichacz considered the group irregularity strength of disconnected graphs in [3].

Theorem 1.4

([3]). Let G be a graph of order n with no component of order less than 3 and with all the bipartite components having both color classes of even order. Let s = n + 1 if n ≡ 2 mod 4 and s = n otherwise. Then: sg(G)=n,ifn1 mod 2,sg(G)=n+1,ifn2 mod 4,sg(G)n+1,ifn0 mod 4.

Moreover, for every integer ts there exists a 𝓖-irregular labeling of G for every Abelian group 𝓖 of order t with at most one involution ι.

Theorem 1.5

([3]). Let G be a graph of order n having neither component of order less than 3 nor a K1, 2u+1 component for any integer u ≥ 1. Then: k(G)=n,ifn1 mod 2,k(G)=n+1,ifn2 mod 4,k(G)n+1,ifn0 mod 4.

Moreover, for every odd integer tk(G) there exists at-irregular labeling of G.

In this paper we give an upper bound for group irregularity strength of all graphs. Moreover, we give the exact values and bounds on sg(G) for disconnected graphs with no star components.

2 Main results

The first natural question is whether the group irregularity strength is finite for arbitrary graph with no components of order less than 3.

Theorem 2.1

Let G be a graph of order n having m components, none of which has order less than 3 and let p be the smallest number greater than 2nm − 1 that has all distinct primes in its factorization. Then sg(G) ≤ p.

Proof

Note that np. Since p has all distinct primes in its factorization, there exists only one (up to isomorphism) Abelian group 𝓖 of order p, namely 𝓖 ≅ ℤp. If all the components of G have order 3 then we are done by Theorem 1.5. Therefore, G has at least one component H such that |V(H)| ≥ 4. Let F be a spanning forest of G. Thus F has nm edges e0, e1, …, enm − 1. Assume without loss of generality that e0E(H). Let f: E(G) → ℤp be defined as follows f(ei)=2i1,for i=1,2,nm1,f(e)=0,for e{e0}E(G)E(F).

We have i=0k12i=2k1<2k for any integer k. Therefore, the maximum weighted degree is smaller than 2nm − 1. Moreover, the unique (additive) decomposition of any natural number into powers of 2 implies that f is 𝓖-irregular. □

A walk in a graph G = (V, E) is a sequence x1, e1, x2, …, xj − 1, ej − 1, vj, where viV, eiE and ei = vi vi+1 for all i. We call the number of vertices of a walk its length. Given any two vertices x1 and x2 that belong to the same connected component of G, there exist walks from x1 to x2. Some of them may consist of even number of vertices (some of them being repetitions). We are going to call them even walks. The walks with odd number of vertices will be called odd walks. We will always choose the shortest even or the shortest odd walk from x1 to x2.

We start with 0 on all the edges of G. Then, in every step we will choose x1 and x2 and add some labels to all the edges of chosen walk from x1 to x2. To be more specific, we will add some element a of the group to the labels of all the edges having odd position on the walk (starting from x1) and −a to the labels of all the edges having even position. It is possible that some labels will be modified more than once, as the walk does not need to be a path. We will denote such situation with ϕe(x1, x2) = a if we label the shortest even walk and ϕo(x1, x2) = a if we label the shortest odd walk. Observe that putting ϕe(x1, x2) = a results in adding a to the weighted degrees of both x1 and x2, while ϕo(x1, x2) = a means adding a to the weighted degree of x1 and −a to the weighted degree of x2. In both cases the operation does not change the weighted degree of any other vertex of the walk. Note that if some component G1 of G is not bipartite, then for any vertices x1, x2G1 there exist both even and odd walks.

We are going to use the following theorem, proved in [17].

Theorem 2.2

([17]). Let s = r1+r2+…+rq be a partition of the positive integer s, where ri ≥ 2 for i = 1, 2, …, q. Let 𝓖 be an Abelian group of order s+1. Then the set 𝓖 ∖ {0} can be partitioned into pairwise disjoint subsets A1, A2, …, Aq such that for every 1 ≤ iq, |Ai| = ri withaAi a = 0 if and only if |𝓖| is odd or 𝓖 contains exactly three involutions.

From the above Theorem 2.2 we easily obtain the following observation:

Observation 2.3

Let s = r1+r2+…+rq be a partition of the positive odd integer s, where ri ≥ 2 for i = 2, 3, …, q. Let 𝓖 be an Abelian group of order s. Then the set 𝓖 can be partitioned into pairwise disjoint subsets A1, A2, …, Aq such that for every 1 ≤ iq, |Ai| = ri withaAi a = 0. □

Using similar method as in the proof of Theorem 1.5 (see Lemma 2.5 from [3], using the result on zero sum partition of cyclic groups from [13]), we can obtain the following lemma.

Lemma 2.4

Let G be a graph of order n having no K1, u components for any integer u ≥ 0. Then for every odd integer tn and for every Abelian group 𝓖 such that |𝓖| = t, there exists a 𝓖-irregular labeling.

Proof

We are going to divide the vertices of G into triples and pairs. Let p1 be the number of bipartite components of G with both color classes odd, p2 with both classes even and p3 with one class odd and one even. Let p4 be the number of remaining components of odd order and p5 - the number of remaining components of even order. The number of triples equals to 2p1+p3+p4. The remaining vertices form the pairs.

By Observation 2.3, the elements of 𝓖 can be partitioned into 2l+1 triples B1, B2, …, B2l+1 and m pairs C1, C2, …, Cm, where l = ⌊(2p1+p3+p4)/2⌋ and m = (t − 6l − 3)/2, such that ∑xBi x = 0 for i = 1, …, 2l+1 and ∑xCj x = 0 for j = 1, …, m. Observe that l ≥ 0, m ≥ 0 and 2l+1 ≥ 2p1+p3+p4.

Let Bi = {ai, bi, ci} for i = 1, 2, …, 2l+1 and let Cj = {dj, − dj} for j = 1, …, m. It is easy to observe that for a given element g ∈ 𝓖 not belonging to any triple, we have (g, −g) = Cj for some j.

Let us start the labeling. For both vertices and labels, we are numbering the pairs and triples consecutively, in the same order as they appear in the labeling algorithm described below, every time using the lowest index that has not been used so far (independently for the lists of couples and triples).

Given any bipartite component G with both color classes even, we divide the vertices of every color class into pairs (xj1,xj2), putting ϕo(xj1,xj2)=dj

for every such pair. We proceed in similar way in the case of all the non-bipartite components of even order, coupling the vertices of every such component in any way.

If both color classes of a bipartite component are of odd order, then they both have at least 3 vertices. We choose three of them, denoted with xj, yj and zj, in one class and another three, xj+1, yj+1 and zj+1, in another one and we put ϕe(xj,zj+1)=aj,ϕe(yj,zj+1)=bj,ϕe(zj,zj+1)=cj,ϕe(xj+1,zj)=aj+1,ϕe(yj+1,zj)=bj+1,ϕe(zj+1,zj)=cj+1.

We proceed with the remaining vertices of these components as in the case when both color classes are even.

In the case of non-bipartite components of odd order we choose three vertices. We put ϕe(xj,zj)=aj,ϕe(yj,zj)=bj,ϕe(zj,zj)=cj.

Finally, for bipartite components of odd order we choose four vertices xj, yj, zj and v (v belongs to the even color class and three other vertices to the odd one). We put ϕe(xj,v)=aj,ϕe(yj,v)=bj,ϕe(zj,v)=cj.

The labeling defined above is 𝓖-irregular. Indeed, in the jth triple of vertices the weights are equal to w(xj) = aj, w(yj) = bj and w(zj) = cj and in the jth pair we have w(xj1)=dj and w(xj2)=dj. Eventually, at least one of the triples of labels remains unused. □

The following theorem easily follows from the above Lemmas 1.3 and 2.4.

Theorem 2.5

Let G be a graph of order n having no K1, u components for any integer u ≥ 0. Then: sg(G)=n,ifn1 mod 2,sg(G)=n+1,ifn2 mod 4,sg(G)n+1,ifn0 mod 4.

We will consider now some families of disconnected graphs of order n ≡ 0 mod 4 for which sg(G) = n.

Proposition 2.6

Let G be a graph of order n ≡ 4 mod 8 with no component of order less than 3 and with all the bipartite components having both color classes of even order. Then sg(G) = n.

Proof

Let 𝓖 be an Abelian group of order n. Since the order of 𝓖 is even there is at least one involution in 𝓖. If there is exactly one involution, then we are done by Theorem 1.4. Thus we can assume that 𝓖 has more than one involution. Observe that n = 22(2α+1) for some integer α, therefore by fundamental theorem of finite Abelian groups we obtain that 𝓖 has exactly three involutions ι1, ι2, ι3.

Let p1 be the number of components of odd order, p2 be the number of components of even order.

Assume first p2 > 0. Then there exists a component H of even order |H| ≥ 4. Note that there exist vertices u, v, x, yV(H) such that there is an odd walk from u to x, an even walk from u to v and an even walk from u to y (if H is bipartite, we take u and x from one color class and v and y from another, what is always possible, since in this case H both color classes have even order). By Theorem 2.2, the set of the elements of 𝓖 ∖ {0} has partition into p1+1 triples B1, B2, …, Bp1+1 and m pairs C1, C2, …, Cm where m = (n − 3p1 − 4)/2 ≥ 0 such that ∑xBi x = 0 for i = 1, …, p1+1 and ∑xCj x = 0 for j = 1, …, m. Let Bp1+1 = {ap1+1, bp1+1, cp1+1}, without loss of generality we can assume that ap1+1 = ι1. Put ϕo(u,x)=ap1+1,ϕe(u,v)=bp1+1,ϕe(u,y)=cp1+1.

Note that we obtain now w(u) = 0, w(x) = −ι1 = ι1, w(v) = bp1+1 and w(y) = cp1+1. We proceed with the remaining vertices in the same way as in the proof of Lemma 2.4 (we divide V(G) ∖ {x, y, u, v} into triples and pairs).

If p2 = 0 then by Theorem 2.2, the set of the elements of 𝓖 ∖ {0} has partition into triples B1, B2, …, Bp1 − 1 and m pairs C1, C2, …, Cm where m = (n − 3p1+2)/2 ≥ 0 such that ∑xBi x = 0 for i = 1, …, p1 − 1 and ∑xCj x = 0 for any j = 1, …, m. We set Bp1 = Cm ∪ {0} and proceed in the same way as in the proof of Lemma 2.4. □

For n ≡ 0 mod 8 we have the following result, unfortunately with a stronger assumption on non-bipartite components:

Theorem 2.7

Let G be a disconnected graph of order n with all components of order divisible by 4 and all the bipartite components having both color classes of even order. Then sg(G) = n.

Proof

Let 𝓖 be an Abelian group of order n. Note that n ≡ 0 mod 4. Since the order of 𝓖 is even there is at least one involution in 𝓖, thus by Theorem 1.4 we can assume that 𝓖 has the set of involutions I = {ι1, ι2, …, ι2p − 1 for some p ≥ 2.

Obviously, I = I* ∪ {0} is a subgroup of 𝓖. Note that Γ = {0, ι1, ι2, ι1+ι2} is a subgroup of I as well as a subgroup of 𝓖. If p = 2 then we define B1 = Γ. If p ≥ 3, then there exists a coset decomposition of I into a1+Γ, a2+Γ, …, a2p − 3+Γ for ajI, j = 1, 2, …, 2p − 3. Set Bj = aj+Γ for j = 1, 2, …, 2p − 3. Obviously, ∑bBj b = 0 and, moreover, for any bBj we have −b = b for j = 1, 2, …, 2p − 3.

Note that the remaining elements of 𝓖 i.e. the elements of 𝓖 ∖ I can be divided into quadruples of four distinct elements Bj={gj1,gj1,gj2,gj2} for j = 2p − 3+1, 2p − 3+2, …, |𝓖|/4, none of which being an involution.

Let Bj={bj1,bj2,bj3,bj4} and bj3{bj1,bj1} for j = 1, 2, …, |𝓖|/4. Let us start the labeling. Given any bipartite component G with both color classes even, we divide the vertices of the component into quadruples (xj1,xj2,xj3,xj4) such that, xj1,xj2 are in the same color class, and xj3,xj4 are in the same color class (possibly the same as xj1,xj2 but not necessarily). We proceed in a similar way in the case of all the non-bipartite components. We are numbering the quadruplets consecutively, starting with 1.

If there is an involution in Bj then set ϕo(xj1,xj2)=bj1,ϕe(xj1,xj3)=bj2,ϕe(xj1,xj4)=bj3.

Observe that in that case w(xj1)=bj1+bj2+bj3=bj4=bj4,w(xj2)=bj1=bj1,w(xj3)=bj2 and w(xj4)=bj3. If there is no involution in Bj then let ϕo(xj1,xj2)=bj1,ϕo(xj3,xj4)=bj3.

Note that we obtain now w(xj1)=w(xj2)=bj1,w(xj3)=w(xj4)=bj3.

The lexicographic product or graph composition GH of graphs G and H is a graph such that the vertex set of GH is the Cartesian product V(G) × V(H) and any two vertices (u, v) and (x, y) are adjacent in GH if and only if either u is adjacent with x in G or u = x and v is adjacent with y in H. Note that GH and HG are not isomorphic in general. One can imagine obtaining GH by blowing up each vertex of G into a copy of H. For instance lK2r, 2rlK2K2r.

One can easily see that if H has no isolated vertices and F is a graph of order divisible by 4, then sg(HF) = |H|⋅|F| by the above Theorem 2.7. Observe also that if H has all components of even order then for G = HK2r we have sg(G) = 2r|H| for any r ≥ 1. One could ask if we need the assumption on the order of components of H. Before we proceed we will need the following result:

Theorem 2.8

([6]). Let s = qr, where r ≥ 3 and 𝓖 be an Abelian group of order s such that the number of involutions in 𝓖 is not one. Then the set 𝓖 can be partitioned into pairwise disjoint subsets A1, A2, …, Aq such that for every 1 ≤ iq, |Ai| = r withaAi a = 0.

Observation 2.9

Let H be a graph of order n with no isolated vertices. If GHK2r for some positive integer r ≥ 2, then sg(G) = 2rn for rn even and sg(G) = 2rn+1 otherwise.

Proof

Obviously, G is a graph of order 2nr with no component of order less than 3 and with all the bipartite components having both color classes of even order. If nr is odd, then 2nr ≡ 2 mod 4, hence sg(G) = 2rn+1 by Theorem 1.4. Therefore, we can assume that 2nr ≡ 0 mod 4. Let 𝓖 be an Abelian group of order 2nr. Since the order of 𝓖 is even there is at least one involution in 𝓖, therefore we can assume that 𝓖 has more than one involution by Theorem 1.4. The set of the elements of 𝓖 has a partition into sets A1, A2, …, An of order 2r such that ∑xAi x = 0 by Theorem 2.8.

Let Ai={ai1,ai2,,ai2r} for i = 1, 2, …, n. Denote the vertices of G corresponding to a vertex xiV(H) by xi1,xi2,,xi2r. Let yNH(xi), then y1N(xij) for j = 1, 2, …, 2r. Set ϕe(xij,y1)=aji for j = 1, 2, …, 2r. One can check that the weighted degrees of all the vertices are distinct. □

Using the same method as in the proof of Observation 2.9 we have the following result.

Observation 2.10

Let H be a graph of order n with no isolated vertices and with all the bipartite components having both color classes of even order. If GHK2r+1 for some positive integer r ≥ 2, then sg(G) = (2r+1)n for n ≢ 2 mod 4 and sg(G) = (2r+1)n+1 otherwise. □

We will finish this section by posting the following conjecture.

Conjecture 2.11

Let G be a graph of order n having no K1, u components for any integer u ≥ 0. Then sg(G) = n if n ≢ 2 (mod 4) and sg(G) = n+1 otherwise.

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About the article

Received: 2017-08-21

Accepted: 2018-02-06

Published Online: 2018-03-02


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 154–160, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0017.

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